What is the extraneous solution to these equations? $\dfrac{x^2}{x + 9} = \dfrac{81}{x + 9}$
Explanation: Multiply both sides by $x + 9$ $ \dfrac{x^2}{x + 9} (x + 9) = \dfrac{81}{x + 9} (x + 9)$ $ x^2 = 81$ Subtract $81$ from both sides: $ x^2 - (81) = 81 - (81)$ $ x^2 - 81 = 0$ Factor the expression: $ (x - 9)(x + 9) = 0$ Therefore $x = 9$ or $x = -9$ At $x = -9$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -9$, it is an extraneous solution.